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Question Number 40787 by rahul 19 last updated on 27/Jul/18

Let I_{1} = \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{\sin x}{x} d x, I_{2} = \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{\sin (\sin x)}{\sin x} d x

, I_{3} = \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{\sin (\tan x)}{\tan x} d x .

P rove that I_{2} > I_{1} > I_{3} .

Commented by tanmay.chaudhury50@gmail.com last updated on 27/Jul/18

f (x) = \frac{s i n x}{x} g (x) = \frac{s i n (s i n x)}{s i n x} h (x) = \frac{s i n (t a n x)}{t a n x}

t o p r o v e I_{2} > I_{1} > I_{3}

w e h a v e t o p r o v e

g (x) > f (x) > h (x) w h e n \frac{Π}{3} > x > \frac{Π}{6}

t a k i n i n g t h e h e l p o f g r a p h i a m t r y i n g t o

e s t a b l i s h \dots l a t e r b y m a t h e m a t i c a l l y

Commented by tanmay.chaudhury50@gmail.com last updated on 27/Jul/18

Commented by tanmay.chaudhury50@gmail.com last updated on 28/Jul/18

i a m a t t a c h i n g s o m e t h i n g r e f r e s h i n g \dots

Commented by tanmay.chaudhury50@gmail.com last updated on 28/Jul/18

Answered by MJS last updated on 27/Jul/18

the integrals represent the areas between

the functions and the x - axis .

so all we have to show is

u \frac{\sin \sin x}{\sin x} ⩾ \frac{\sin x}{x} ⩽ \frac{\sin \tan x}{\tan x} \forall x \in [\frac{π}{6}; \frac{π}{3}]

for x = \frac{π}{6} we get

2 \sin \frac{1}{2} ⩾ \frac{3}{π} ⩾ \sqrt{3} \sin \frac{\sqrt{3}}{3}

. 958 \dots ⩾ . 954 \dots ⩾ . 945 \dots

for x = \frac{π}{3} we get

\frac{2 \sqrt{3}}{3} \sin \frac{\sqrt{3}}{2} ⩾ \frac{3 \sqrt{3}}{2 π} ⩾ \frac{\sqrt{3}}{3} \sin \sqrt{3}

. 87 \dots ⩾ . 82 \dots ⩾ . 56 \dots

we have to show that the functions are

falling within the given interval (they {don}^{'} t

have any “ {jumps}^{″} or “ {holes}^{″})

x \in [\frac{π}{6}; \frac{π}{3}] \Rightarrow

\Rightarrow \sin x \in [\frac{1}{2}; \frac{\sqrt{3}}{2}] \Rightarrow \sin \sin x is increasing

\Rightarrow \tan x \in [\frac{\sqrt{3}}{3}; \sqrt{3}] \Rightarrow \sin \tan x is increasing

both without any salience

the denominators x, \sin x and \tan x have no

zeros in the given intervall

\Rightarrow try to find the amount of increasement

in the interval

(\int f^{'} (x) d x = f (x))

{[\frac{\sin \sin x}{\sin x}]}_{\frac{π}{6}}^{\frac{π}{3}} \approx - . 079

{[\frac{\sin x}{x}]}_{\frac{π}{6}}^{\frac{π}{3}} \approx - . 128

{[\frac{\sin \tan x}{\tan x}]}_{\frac{π}{6}}^{\frac{π}{3}} \approx - . 376

so those who start at a lower value loose

more along the way \Rightarrow this should be proven

or simply plot them; -)

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